Do projection operators preserve the non-negativity of the trace?

Question

I have a question involving projection operators and the trace of a positive operator.

More specifically, let $V$ be a complex inner product space, $A$ be a positive operator on $V$ and $P$ be a projection operator on $V$. What I would like to know is if it is always the case that $\text{tr}(PA)\geq0$.

I tried to express $PA$ as the composition of a linear operator and its adjoint, but to no avail. Also, I do know that if a counterexample exists, the dimension of $V$ must be at least 3. As such, I would like to know how would one go about proving or disproving the statement.

Answer 1

Hint: ${\rm tr}(P A) = {\rm tr}(P^2 A) = {\rm tr}(PAP)$.

Answer 2

More generally, if $A$ and $B$ are positive, then $\mathrm{tr}(BA)=\mathrm{tr}\left({\sqrt BA\sqrt B}\right)\geq 0$.

Related:

If $a\ge 0$ and $b\ge 0$, then $\sigma(ab)\subset\mathbb{R}^+$.